Figuring proportions?

[Woodtroll]

Folks,

Algebra having been years (and years) behind me, I cannot for the life of me remember how to figure proportions of alloy components. For example, say I have some pure lead, and some 50/50 bar solder, and I want to make a nice 20:1 blackpowder alloy. As I add tin to the mix, I am also adding lead, increasing that proportion as I increase the tin. Can someone refresh me on the math involved for this. How do I figure how much 50/50 to add to have a final ratio of 20:1?

How about 3-part alloys? Say I have some 94/6 lead/antimony alloy, and want to add 50/50 bar solder to make Lyman #2 or some other alloy? What is the mathematical (or algebraic) means to figure this?

Thanks for your help; the older I get, the more I seem to leave behind!

Y'all take care! Regan

[Bob Krack]

Folks,
Algebra having been years (and years) behind me, I cannot for the life of me remember how to figure proportions of alloy components. For example, say I have some pure lead, and some 50/50 bar solder, and I want to make a nice 20:1 blackpowder alloy. As I add tin to the mix, I am also adding lead, increasing that proportion as I increase the tin. Can someone refresh me on the math involved for this. How do I figure how much 50/50 to add to have a final ratio of 20:1?

How about 3-part alloys? Say I have some 94/6 lead/antimony alloy, and want to add 50/50 bar solder to make Lyman #2 or some other alloy? What is the mathematical (or algebraic) means to figure this?

I think ya got a pretty good question and there are many here that will probably answer shortly.

I always did it the slow "stick" method.

100 lbs of lead, 100 lbs of 50/50 = 150 lead and 50 tin. (3 to 1)Hmmmmm...
let's try 200 lbs lead, 100 lbs 50/50 = 250 lead 50 tin (5 to 1) and keep trying til I get it right.

The most confusing to me is the term 10:1 which I take to mean 10 parts lead and 1 part tin, whereas I have seen 10:1 as meaning 10 pounds total of which 9 parts are lead and 1 part is tin.

How's that for muddy'n the water? (They don't call me VillageIdjit for nothin!)

Vic

[Linstrum]

Yup, VillageIdjit has the right attitude as far as getting things done in the real world. I taught algebra and calculus in college, but what he says is what I also do because IT IS FASTER! I just hate the math programs nowadays because they don't teach real world situations where getting the job done close but not exact is just plain good enough.

You can understand what to do if you break up the math problem into separate steps and then define each component as to how much it weighs and then just add them all up. Let's say you have a pound of 96% lead and 4% antimony alloy. That is 0.96 pounds lead and 0.04 pounds antimony. Add them up and you have 0.96 pounds + 0.04 pounds = 1 pound total weight.


Simple so far.


Okay, we still have 0.96 pounds lead and 0.04 pounds antimony. If we add a pound of 50/50 solder to it we now have 2 pounds total weight. Breaking that down into its separate parts we have 0.96 pounds lead + 0.04 pounds antimony + 0.5 pounds lead + 0.5 pounds tin. Now add them up but keep each kind of component separate as follows:


0.96 pounds lead + 0.5 pounds lead


0.04 pounds antimony


0.5 pounds tin


Totaling them up we have 1.46 pounds lead in the mix, 0.4 pounds antimony, and 0.5 pounds tin. Together they weight 2 pounds.


The next step is to mathematically find out what the percentages are of each one, which is as follows:


1.46 divided by 2 = 0.73 or 73% lead


0.04 divided by 2 = 0.02 or 2% antimony


0.5 divided by 2 = 0.25 or 25% tin


Adding those up to check your math they should equal 100%.


73 + 2 + 25 = 100


It checks out okay.


Going backwards is a lot harder. I seem to recall the subject of what on earth quadratic equations are good for coming up about a week ago, and one solution for this requires the use of the quadratic function to solve. It is far easier and quicker to guesstimate and approximate to get close than to sit down and derive the formula to solve directly using algebra.


Let's say somebody gave you some of the 96%/4% mix and we need to find out how many pounds of 50/50 solder we need to add to it to make some boolit alloy that is 4% tin. We have a pound of alloy and we want to add some 50/50 to it so that the end result is 4% tin. Altogether it will have to be 8% 50/50 solder because half of 8% is 4% that we are after. If we add 0.08 pounds of 50/50 to it the end result is 0.04 pounds tin divided by 1.08 pounds new alloy or 0.0373% tin, which is close to 4% If we add 0.0864 pounds 50/50 to it we get 3.98%, getting closer. It is easier to keep track of how much we estimate we put in then do the addition of the individual components like in the examples above. Calculus can be used as well, but again doing the by guess and by golly method is still easier and is ACCURATE ENOUGH that way, you can't tell the difference between 3.98% tin from 4% tin alloy! Sitting down and deriving the formula needed for integral calculus to get the exact answer I can do but I don't have time to derive the equation. Yeah, it is a cop out, I last used integral calculus 5 years ago and I'm a little rusty.


Have fun!


rl405

[Tom Myers]

You can denote a tin lead alloy in 2 different ways.
A 1 part tin and 19 part lead alloy can be referred to as 1 in 20 and written as 1 / 20 or 1:19 which would contain 5% of tin in the alloy.

A 1 part tin and 20 part lead alloy can be referred to as 1 and 20 and written as 1:20 or 1 / 21 which would contain 4.762% tin in the alloy.


To calculate the amount of the Add Alloy to mix in with the Old Alloy in order to arrive at a certain percentage of Tin in the New Alloy:

Let X = the Weight of the Add Alloy which will be a 50% tin and 50% lead alloy.

Let the New Alloy be a 1 in 20 or 1 / 20 Alloy which will contain 5% Tin.

Let The Old Alloy be pure lead which will contain 0% Tin.

OldAlloyWt * ( 1 - %Tin in OldAlloy / 100) + X * ( 1 - % Tin in AddAloy / 100 ) = ( OldAlloyWt + X ) * ( 1 - %Tin in NewAlloy / 100 )

%Tin in OldAlloy = 0
%Tin in AddAlloy = 50
%Tin in NewAlloy = 5

10 pounds of Old Alloy

10 * ( 1 - 0 / 100 ) + X * ( 1 - 50 / 100 ) = (10 + X ) * ( 1 - 5 / 100 )

10 * ( 1- 0 ) + X * ( 1 - 0.5 ) = ( 10 + X ) * ( 1 - 0.05)

10 * 1 + X * 0.5 = ( 10 + X ) * 0.95

10 + X - 0.5*X = 9.5 + 0.95*X

10 - 9.5 = 0.95*X - .5*X

0.5 = 0.45*X

X = 0.5 / 0.45

X = 1.1111 pounds of Add Alloy to add to the 10 pounds of Old Alloy

This adds up to 11.111 lbs of alloy of which 0.55555 lbs is tin

0.55555 / 11.1111 = 0.05 or 5% Tin or a 1 in 20 Tin Lead Alloy

If you have wheelweights and want to add enough 60:40 Tin Lead Solder to arrive at a 5% tin alloy, it goes like this

Wheelweights = 95% Lead, 4.5% Antimony and 0.5% Tin

10 lb Wheelweights = 0.5% Tin
New Alloy = 5% Tin
Solder = 60% Tin

10 * ( 1 - 0.5 / 100 ) + X * ( 1 - 60 / 100 ) = (10 + X ) * ( 1 - 5 / 100 )
10 * 0.995 + X - X * 0.6 = ( 10 + X ) * 0.95
9.95 + 0.4*X = 9.5 + 0.95*X
9.95 - 9.5 = 0.95*X - 0.4*X
0.45 = 0.55*X
X = 0.45 / 0.55
X = 0.818181 = lbs of 60:40 solder to add to 10 lbs of wheel weights to have a 5%tin alloy
This New Alloy contains 4.15% Antimony 5% Tin and 90.85% Lead


Hope this helps
Tom Myers
Precision Ballistics and Records (http://www.tmtpages.com)

[garandsrus]

Woodtroll,

If you have Excel, there is a current thread that has an alloy calculation spreadsheet attached and there will be another spreadsheet attached shortly.

John

[Lloyd Smale]

to make 20 to 1 add 10 lbs of your 5050 solder to 95lbs of pure or 5 lbs of it to 50lbs. close enough. to make #2 out of your 94/6 add 10pecent by weight of you 5050 and youll be close enough again. thats more then you really need but true #2 has a bit more tin then nessisary in it. 6 percent would give you a good casting alloy.

[Woodtroll]

Thanks, guys, for all the help. I do everything now by the reason-it-out method most of you posted, but thought that maybe there was an easier way. The spreadsheet in the other thread is helpful, too.

For what it is worth, I think the standard nomenclature of the lead/tin alloys when blackpowder was king was to denote the "parts" of each, similar to recipes of the time. Thus, 20:1 is 20 parts lead, 1 part tin, for 21 total parts. Most BP shooters, I believe, put the lead first in the ratio.

I had never paid attention to the difference between 1/20 and 1:20 (or, as I prefer, 20:1) until Tom pointed it out, and of course he is right; that explains the difference that VillageIdjit made mention of. To confuse things even more, at least one BPCR supplier advertises "20/1" alloy. Now that Tom has shown me the difference, I find it funny, but before he pointed out my ignorance I (thought I) knew what he meant.

Great discussion, folks, and I sure do appreciate the help!

Y'all take care! Regan

[felix]

Most professional warehouses use the first number as the largest component, and progress downwards. The problem exists always: what is the order of the components in nomenclature? You just have to know in advance of ordering anything. For example, what we call 60/40 radiator solder. It is actually closer to 63/37. But, what is the 63? TIN, is the answer, followed by lead. How would you know? By calling the warehouse; the only way for sure. ... felix

[Ken O]

Does this mean I've been doing it wrong? For 3:1 (lead/lino) I would put 30 pounds of lead and 10 pounds of lino. I went to public high school, so go easy on me.

[Bob Krack]

Does this mean I've been doing it wrong? For 3:1 (lead/lino) I would put 30 pounds of lead and 10 pounds of lino. I went to public high school, so go easy on me.

Where I grew up, the ":" in the "3:1" meant "to" as in 3 parts to 1 part.

Not making any waves here, I've seen it both ways. I've seen LASC articles showing the proportion of tin in 10:1 as 91% lead and 9% tin and when I talked to Lee and RCBS service techs they said "9 parts lead and 1 part tin" as the ratio they designed their moulds to cast rated weights.

Long ago and far away as a weldor, the specifications called for a "1 inch in 4 inch" stitch weld and the spec usually meant to weld one inch and skip 4 inches. I always felt that the terminology meant to weld one inch and skip three inches. (Blueprint specs of 1 in 4 meant weld 1 skip 3). Shop and field supervisors usually meant to weld 1 and skip 4. Go figger.

I think I am happy with LASC nomenclature, but I can live with any other.

I prefer 92% lead, 6% antimony, and 2% tin with a trace of arsenic if I could choose it all myself. I'm happy with wheel weights as the easiest to come up with and to work with.

Vic

[runfiverun]

i just base the #'s to equal 100 lbs and if i only need 20#'s i divide by 5.
if you keep doing it this way you will have the same answer each time, and you will have worked your loads out for your alloy.
the #'s don't matter the consistent results do.
if it is 5% or 4.74% don't matter. what does is that you are close enough to either 5 or 4.74 each time.
to not tell the difference on target.

[Woodtroll]

Runfiverun,

I agree with you one hundred percent! When I find an alloy that works with a prticular load or gun, I'd like to be able to reproduce it consistently, but I don't particularly care if it is 20:1 or 19.5:1, as long as it works. The trick is being able to reproduce a particular alloy given more than one starting alloy (I'm a scrounger!), and that is one reason I asked this question to begin with. Thanks for making this point better than I probably would have.

Regan

[John Boy]

Gents ... you would be wise to copy Linstrum's and Tom Myers calculations into your references - for those of us that are mentally challanged!

Thanks Gentlemen for the posts :drinks:

[bobk]

All,
Hate to be contrary, but if you have 19 lb. lead and 1 lb. tin, versus 20 lb. lead and 1 lb. tin, what possible difference will it make? As a practical matter, does anyone actually weigh their ingots, or do we assume that they're a certain weight? Precise alloy mix is maybe the least important factor in cast boolit accuracy.
Bob K

[garandsrus]

Bob,

If I am trying to make a specific (or close to it) alloy, I will weigh the ingots of the various components. If not, WW with a little tin works great :)

John

[Linstrum]

I do weigh alloy components, all the way from 100 pounds wheel weights down to grains of phosphorus-copper alloy for hardening small ounce-size batches of special alloy for hunting boolits to take the place of EXPENSIVE Barnes-X copper bullets. You can get Barnes-X performance from cast alloy by adding about 5% phoscopper brazing alloy to wheel weight alloy.

But my point is that I do weigh things carefully and keep records. Bathroom scales nowadays are pretty accurate and work great for making big batches, just calibrate them first, a gallon jug of water weighs in at exactly 8-1/3 pounds (the old saying "a pint a pound the world 'round" IS WRONG WRONG WRONG!!!), so 3 gallons of water weighs exactly 25 pounds, 12 gallons of water weighs 100 pounds.

But as in my example above from a few days ago, keep your measuring accuracy in perspective, 100 pounds of wheel weight alloy weighed to the nearest grain is stupid as well as time consuming (unless you have a laboratory scale), just getting it down close to within a pound or so of 100 pounds is good enough and you will be okay.


rl409

[bobk]

Linstrum,
Ah, you want to expand (pun intended!) on this? Seriously, I am interested because I would like to be able to make an alloy that would be good for solids. How is such a thing affected by water dropping, or oven heat treating? Or would I be better off just adding lead shot (for the arsenic) to WW and water dropping?
Bob K

[Lloyd Smale]

i dont worry about exact measurment either but when i do a batch i do a big one so that the bullets are at least consistant for a while.

[Linstrum]

Hey, bobk, glad to tell you more about my phosphorus-copper-wheel wight alloy boolits.

I am going to move it over to SPECIAL PROJECTS, otherwise it will get lost here in this thread since it is a bit different subject.


rl411

[bobk]

Linstrum,
I'll be looking for it!
Bob K